# Math Help - Cylindrical coordinates to cartesian coordinates

1. ## Cylindrical coordinates to cartesian coordinates

Hi there. Hi have in cylindrical coordinates that $\theta=\displaystyle\frac{\pi}{3}$, and I must make the graph, and take it into cartesian coordinates. How should I do?

I've tried this way:

$\begin{Bmatrix}x=r\cos\displaystyle\frac{\pi}{3}\\ y=r\sin\displaystyle\frac{\pi}{3} \\z=z\end{matrix}\Rightarrow{\begin{Bmatrix}x=\dis playstyle\frac{r}{2}\\y={r\displaystyle\frac{\sqrt[ ]{3}}{2} \\z=z\end{matrix}}$

I think its a semi-plane parallel to the line: $2\displaystyle\frac{y}{\sqrt[ ]{3}}-2x=0$. I thought of working geometrically with it, taking another point. Or taking three points, but I think its probably easier someway, just from the equations system. I don't know how to take x and y, to make them a function of z.

Bye there!

2. Originally Posted by Ulysses
Hi there. Hi have in cylindrical coordinates that $\theta=\displaystyle\frac{\pi}{3}$, and I must make the graph, and take it into cartesian coordinates. How should I do?

I've tried this way:

$\begin{Bmatrix}x=r\cos\displaystyle\frac{\pi}{3}\\ y=r\sin\displaystyle\frac{\pi}{3} \\z=z\end{matrix}\Rightarrow{\begin{Bmatrix}x=\dis playstyle\frac{r}{2}\\y={r\displaystyle\frac{\sqrt[ ]{3}}{2} \\z=z\end{matrix}}$

I think its a semi-plane parallel to the line: $2\displaystyle\frac{y}{\sqrt[ ]{3}}-2x=0$. I thought of working geometrically with it, taking another point. Or taking three points, but I think its probably easier someway, just from the equations system. I don't know how to take x and y, to make them a function of z.

Bye there!
$\theta = \frac{ \pi }{3}$ corresponds to special triangles. Special right triangles - Wikipedia, the free encyclopedia

3. How do I apply that on this problem?

Thanks btw.

4. Hello, Ulysses!

Your work is correct . . . just put it together.

$\text{In cylindrical coordinates: }\;\theta\:=\: \frac{\pi}{3}$

$\text{I must take it into cartesian coordinates and graph.}$

$\text{I've tried this way:}$

$\begin{Bmatrix}x&=&r\cos\frac{\pi}{3}\\ \\[-3mm] y&=&r\sin\frac{\pi}{3} \\ \\[-3mm] z&=&z\end{matrix}\quad\Rightarrow\quad \begin{Bmatrix}x&=& \frac{1}{2}r & [1] \\ \\[-3mm] y&=&{\frac{\sqrt{3}}{2}\,\!r & [2] \\ \\[-3mm] z&=&z\end{matrix}}$

$\text{I think its a semi-plane parallel to the line: }\:\frac{2}{\sqrt{3}}y-2x\:=\:0$
Um, it's the entire plane . . .

$\begin{array}{cccccc}\text{From [1], we have: }& r &=& 2x \\ \\[-3mm]
\text{From [2]. we have:} & r &=& \frac{2}{\sqrt{3}}y \end{array}$

$\text{Hence: }\;\frac{2}{\sqrt{3}}y \:=\:2x \quad\Rightarrow\quad y \:=\:\sqrt{3}\,x$

$\text{This is a vertical plane passing through the }z\text{-axis,}$
. . $\text{making a }60^o\text{ angle with the positive }x\text{-axis.}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

By the way . . .

. . \frac{2}{3} produces: . $\frac{2}{3}$

. . \dfrac{2}{3} produces: . $\dfrac{2}{3}$

5. Thank you very much Soroban.

So, I must use the fact of that its a plane parallel to that line, and work geometrically to get it into a function of z, right?

6. It'll be a semi-plane if you're prohibiting negative r-values, and a full plane if you're allowing negative r-values. Polar coordinates usually allow negative r-values. By extension, then, I would think cylindrical coordinates would usually allow negative r-values.

7. Originally Posted by Soroban
Hello, Ulysses!

Your work is correct . . . just put it together.

$\begin{array}{cccccc}\text{From [1], we have: }& r &=& 2x \\ \\[-3mm]
\text{From [2]. we have:} & r &=& \frac{2}{\sqrt{3}}y \end{array}$

$\text{Hence: }\;\frac{2}{\sqrt{3}}y \:=\:2x \quad\Rightarrow\quad y \:=\:\sqrt{3}\,x$

$\text{This is a vertical plane passing through the }z\text{-axis,}$
. . $\text{making a }60^o\text{ angle with the positive }x\text{-axis.}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

By the way . . .

. . \frac{2}{3} produces: . $\frac{2}{3}$

. . \dfrac{2}{3} produces: . $\dfrac{2}{3}$
Note how this corresponds to the special triangles noted above.

The angle $\frac{ \pi }{3}$ is given by the triangle with the sides $\sqrt{3}$, and $1$ with the hypotenuse $2$.

Rise over run gives $\frac{y}{x} = \frac{ \sqrt{3} }{1} \to y = \sqrt{3}x$
This does not have a z component.